Trace Sigma-Algebra of Generated Sigma-Algebra
Theorem
Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $A \subseteq X$ be a subset of $X$.
Then the following equality holds:
- $A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
- $\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
- $\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$
- $A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$
- $A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$
Proof
By definition of generated $\sigma$-algebra:
- $\GG \subseteq \map \sigma \GG$
whence from Set Intersection Preserves Subsets:
- $A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of generated $\sigma$-algebra:
- $\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$
For the reverse inclusion, define $\Sigma$ by:
- $\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$
We will show that $\Sigma$ is a $\sigma$-algebra on $X$.
Since $A \in \map \sigma {A \cap \GG}$:
- $A \cap X = A \in \map \sigma {A \cap \GG}$
and therefore $X \in \Sigma$.
Suppose that $E \in \Sigma$.
Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset:
- $\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$
Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$:
- $A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.
Then by Intersection Distributes over Union:
- $\ds \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$
The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$.
Therefore, $\Sigma$ is a $\sigma$-algebra.
It is also apparent that $\GG \subseteq \Sigma$ since:
- $A \cap \GG \subseteq \map \sigma {A \cap \GG}$
by definition of generated $\sigma$-algebra.
Thus, as $\Sigma$ is a $\sigma$-algebra:
- $\map \sigma \GG \subseteq \Sigma$
and therefore:
- $A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$
Hence the result, by definition of set equality.
$\blacksquare$